Question

(4x+3y^2)dx+2xy*dy=0 by using integrating fector

Answer 1

`\underbrace{(4x+3y^2)}_M\,\mathrm dx+\underbrace{2xy}_N\,\mathrm dy=0`

The ODE is exact if
`(\partial M)/(\partial y)=(\partial N)/(\partial x)`.


`M_y=6y`

`N_x=2y`

This is not the case, so look for an integrating factor
`\mu(x)` such that


`\frac\partial{\partial y}\mu M=\frac\partial{\partial x}\mu N`

Since
`\mu` is a function of
`x` only, you have


`\mu M_y=\mu'N+\mu N_x\implies\frac{\mu'}\mu=\frac{M_y-N_x}N\implies\mu=\exp\left(\displaystyle\int\frac{M_y-N_x}N\,\mathrm dx\right)`

So, the integrating factor is


`\mu=\exp\left(\displaystyle\int(6y-2y)/(2xy)\,\mathrm dx\right)=\exp\left(2\int\frac{\mathrm dx}x\right)=x^2`

Now the ODE can be modified as


`\underbrace{(4x^3+3x^2y^2)}_(M^*)\,\mathrm dx+\underbrace{2x^3y}_(N^*)\,\mathrm dy=0`

Check for exactness:


`{M^*}_y=6x^2y`

`{N^*}_x=6x^2y`

so the modified ODE is indeed exact.

Now, you're looking for a solution of the form
`\Psi(x,y)=C`, since differentiating via the chain rule yields


`(\mathrm d)/(\mathrm dx)\Psi(x,y)=\Psi_x+\Psi_y(\mathrm dy)/(\mathrm dx)=0`

Matching up components, you would have


`\Psi_x=M^*=4x^3+3x^2y^2`

`\displaystyle\int\Psi_x\,\mathrm dx=\int(4x^3+3x^2y^2)\,\mathrm dx`

`\Psi=x^4+x^3y^2+f(y)`

Differentiate this with respect to
`y` to get


`\Psi_y=2x^3y+f'(y)=2x^3y=N^*`

`f'(y)=0\implies f(y)=C_1`

So the solution here is


`\Psi(x,y)=x^4+x^3y^2+C_1=C\implies x^4+x^3y^2=C`

Just for a final check, take the derivative to get back the original ODE:


`(\mathrm d)/(\mathrm dx)[x^4+x^3y^2]=(\mathrm d)/(\mathrm dx)C`

`4x^3+3x^2y^2+2x^3y(\mathrm dy)/(\mathrm dx)=0`

`4x+3y^2+2xy(\mathrm dy)/(\mathrm dx)=0`

`(4x+3y^2)\,\mathrm dx+2xy\,\mathrm dy=0`

so the solution is correct.